Find $y'$ for $\ln(x+y)=\arctan(xy)$
Here is my attempt at a solution. Is this correct? Any hints or advice would be appreciated.
[Math] Find $y’$ for $\ln(x+y)=\arctan(xy)$
calculusderivativesimplicit-differentiationsolution-verification
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Best Answer
whilst your answer cannot be improved on, it may be of interest to look at a parametric approach. if we set: $\xi $ to be the common value of $\log(x+y)$ and $\arctan(xy)$ then $x$ and $y$ are the two roots of: $$ z^2 -e^{\xi}z + \tan \xi =0 $$ differentiating with respect to $\xi$ gives: $$ 2zz' - e^{\xi}(z+z') +\sec^2 \xi =0 $$ or $$ z' (2z-x-y) = (x+y)z - (1+x^2y^2) $$ if we substitute the roots $x$ and $y$ for $z$ we obtain: $$ z_y' (y-x) = (x+y)y-(1+x^2y^2) \\ z_x' (x-y) = (x+y)x-(1+x^2y^2) $$ hence: $$ \frac{dy}{dx} = \frac{z_y'}{z_x'} = - \frac{(x+y)y-(1+x^2y^2)}{(x+y)x-(1+x^2y^2)} $$